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2 years ago

A(n) _______________ can be formed by linking together several monosaccharides via glycosidic bonds.

Chemistry

1 answer:

mario62 [17]2 years ago

7 0

Answer:

A polysaccharide (n) can be formed by linking several monosaccharides through glycosidic linkages.

Explanation:

Polysaccharides are carbohydrates or complex carbohydrates, where monosaccharides join with glucosidic bonds to form a more complex structure that would be the polysaccharide.

An example of a polysaccharide is starch, or glycogen.

Starch is found in many foods such as potatoes or rice, and glycogen is a form of energy reserve of our organism housed in muscles and liver to fulfill locomotion, physical activity, and other activities that consist of glycolysis.

Polysaccharides are degraded in our body by different stages, and several enzymes unlike monosoccharides or disaccharides, since they have more unions and a more complex structure to disarm in our body and thus assimilate it.

Polysaccharides are also part of animal structures, such as insect shells or nutritional sources, among others.

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A 0.4657 g sample of a pure soluble bromide compound is dissolved in water, and all of the bromide ion is precipitated as AgBr b

Yanka [14]

Answer:

The mass percentage of bromine in the original compound is 81,12%

Explanation:

Step 1: Calculate moles AgBr

moles AgBr = mass AgBr / molar mass AgBr

= 0.8878 g / 187.77 g/mol

= 0.00472812 moles AgBr

⇒

Since 1 mol AgBr contains 1 mol Br-

Then the amount of moles Br- in the original sample must also have been 0.00472812 moles

Step 2: Calculating mass Br-

mass Br- = molar mass Br x moles Br-

= 79.904 g/mol x 0.00472812 mol

= 0.377796 g Br-

⇒

There were 0.377796 g Br- in the original sample

Step 3: Calculating mass percentage Br-

⇒mass percentage = actual mass Br- / total mass x 100%

% mass Br = 0.377796 g / 0.4657 g x 100 %

= 81.12%

7 0

2 years ago

If 36.9 mL of B2H6 reacted with excess oxygen gas, determine the actual yield of B2O3 if the percent yield of B2O3 was 75.7%. (T

zhuklara [117]

Answer: The actual yield of $B_2O_3$ is 60.0 g

Explanation:-

The balanced chemical reaction :

$B_2H_6(l)+3O_2(g)\rightarrow B_2O_3(s)+3H_2O(l)$

Mass of $B_2H_6$ = $Density\times Volume=1.131g/ml\times 36.9ml=41.7g$

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} B_2H_6=\frac{41.7g}{27.668g/mol}=1.51moles$

According to stoichiometry:

1 mole of $B_2H_6$ gives = 1 mole of $B_2O_3$

1.51 moles of $B_2H_6$ gives = $\frac{1}{1}\times 1.51=1.51$ moles of $B_2O_3$

Theoretical yield of $B_2O_3=moles\times {Molar mass}}=1.14mol\times 69.62g/mol=79.3g$

Percent yield of $B_2O_3$ = $75.7\%$

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

$75.7\%=\frac{\text{Actual yield}}{79.3}\times 100$

${\text{Actual yield}}=60.0g$

Thus the actual yield of $B_2O_3$ is 60.0 g

7 0

2 years ago

A buffer is prepared by mixing hypochlorous acid ( HClO ) and sodium hypochlorite ( NaClO ) . If a strong base, such as NaOH , i

Dafna1 [17]

Answer: The balanced chemical equation is written below.

Explanation:

We are given:

A weak acid that is hypochlorous acid (HClO) and basic salt that is sodium hypochlorite (NaClO)

When a strong base is added to the buffer, the hydroxide ion will be neutralized by hydrogen ions from the acid.

So, the buffer component that neutralizes the additional hydroxide ions in the solution is HClO

The chemical equation for the neutralization of hydroxide ion with acid follows:

$HClO+OH^-\rightarrow H_2O+ClO^-$

Hence, the balanced chemical equation is written below.

4 0

2 years ago

HELP ASAP!!!!!!

blondinia [14]

2, 4, 1

Explanation:

We have the following chemical reaction:

Ag₂O → Ag + O₂

To balance the chemical equation the number of atoms of each element entering the reaction have to be equal to the number of atoms of each element leaving the reaction, in order to conserve the mass.

So the balanced chemical equation is:

2 Ag₂O → 4 Ag + O₂

Learn more about:

balancing chemical equations

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#learnwithBrainly

6 0

1 year ago

A piece of iron metal is heated to 155 degrees C and placed into a calorimeter that contains 50.0 mL of water at 18.7 degrees C.

Korvikt [17]

Answer:

D = 28.2g

Explanation:

Initial temperature of metal (T1) = 155°C

Initial Temperature of calorimeter (T2) = 18.7°C

Final temperature of solution (T3) = 26.4°C

Specific heat capacity of water (C2) = 4.184J/g°C

Specific heat capacity of metal (C1) = 0.444J/g°C

Volume of water = 50.0mL

Assuming no heat loss

Heat energy lost by metal = heat energy gain by water + calorimeter

Heat energy (Q) = MC∇T

M = mass

C = specific heat capacity

∇T = change in temperature

Mass of metal = M1

Mass of water = M2

Density = mass / volume

Mass = density * volume

Density of water = 1g/mL

Mass(M2) = 1 * 50

Mass = 50g

Heat loss by the metal = heat gain by water + calorimeter

M1C1(T1 - T3) = M2C2(T3 - T2)

M1 * 0.444 * (155 - 26.4) = 50 * 4.184 * (26.4 - 18.7)

0.444M1 * 128.6 = 209.2 * 7.7

57.0984M1 = 1610.84

M1 = 1610.84 / 57.0984

M1 = 28.21g

The mass of the metal is 28.21g

3 0

1 year ago